91Ó°ÊÓ

Consider the given vector $\stackrel{→}{v}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$and localid="1659429971902" $\stackrel{→}{x}=\left[\begin{array}{c}{\stackrel{→}{x}}_1\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right]$.

If the vector$\stackrel{\mathbf{→}}{\mathbf{x}}$is perpendicular then .

As$\stackrel{→}{v}$is perpendicular to $\stackrel{→}{x}$, by the definition of orthogonal $\stackrel{→}{x}.\stackrel{→}{v}=0$.

Simplify the equation$\stackrel{→}{x}.\stackrel{→}{v}=0$as follows.

$$\begin{gathered} {\stackrel{→}{x}}_1.{\stackrel{→}{v}}_1=0 \\ \left[{\stackrel{→}{x}}_1,{\stackrel{→}{x}}_2,{\stackrel{→}{x}}_3,{\stackrel{→}{x}}_4\right]\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]=0 \\ {\stackrel{→}{x}}_1+2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4=0 \\ {\stackrel{→}{x}}_1=0\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\} \end{gathered}$$

Substitute the value $-\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\}$for ${\stackrel{→}{x}}_1$in the equation $\stackrel{→}{x}=\left[\begin{array}{c}{\stackrel{→}{x}}_1\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right]$as follows.

$$\begin{gathered} \stackrel{→}{x}=\left[\begin{array}{c}{\stackrel{→}{x}}_1\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right] \\ \stackrel{→}{x}=\left[\begin{array}{c}-\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\}\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right] \end{gathered}$$

Therefore, any vector in the form $\stackrel{→}{x}=\left[\begin{array}{c}-\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\}\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right]$is perpendicular to $\stackrel{→}{v}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]=$.

Simplify the equation $\stackrel{→}{x}=\left[\begin{array}{c}-\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\}\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right]$as follows.

$$\begin{gathered} \stackrel{→}{x}=\left[\begin{array}{c}-\left\{2{\stackrel{→}{x}}_2+3{\stackrel{→}{x}}_3+4{\stackrel{→}{x}}_4\right\}\\ {\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_3\\ {\stackrel{→}{x}}_4\end{array}\right] \\ =\left[\begin{array}{c}-2{\stackrel{→}{x}}_2\\ {\stackrel{→}{x}}_2\\ 0\\ 0\end{array}\right]+\left[\begin{array}{c}-3{\stackrel{→}{x}}_2\\ 0\\ {\stackrel{→}{x}}_3\\ 0\end{array}\right]+\left[\begin{array}{c}-4{\stackrel{→}{x}}_4\\ 0\\ 0\\ {\stackrel{→}{x}}_4\end{array}\right] \\ ={\stackrel{→}{x}}_2\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right]+{\stackrel{→}{x}}_3\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\end{array}\right]+{\stackrel{→}{x}}_4\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right] \end{gathered}$$

Therefore, the vector $\stackrel{→}{x}∈span\left\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$.

Hence, the vector$\stackrel{→}{x}$ is spanned by$\left\{\left(\begin{array}{cccc}-2& 1& 0& 0\end{array}\right),\left(\begin{array}{cccc}-3& 0& 1& 0\end{array}\right),\left(\begin{array}{cccc}-4& 0& 0& 1\end{array}\right)\right\}$ which is perpendicular to $\stackrel{→}{v}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$.